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    Ok so first we need to consider the power output for a lower voltage, using P=\frac{V^2}{R} and assuming a constant resistance (for now) we know that the power is reduced (bear in mind that this step is purely to help your understanding and isn't strictly physical). Since there is lower power there is less heat produced. We know that if there if the filament is cooler, it's resistance will be reduced resulting in a higher current. Since the resistance doesn't increase linearly and applying V=IR, if R is less than the original R then the current for the same voltage will be higher. We can apply this to show that the power is higher than 9W.
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    With filamnet lamps, as you increase the voltage, the resistance increases as there are more collisions between electrons and the ions in the metal. This is because filament lamps are non-ohmic.

    So, if you reduce the voltage, there are less collisions so the resistance and hence the temperature will be lower - and less energy dissipated as heat. Therefore, the power rating will be greater.
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    (Original post by voltz)
    and less energy dissipated as heat. Therefore, the power rating will be greater.
    Power is energy dissipated over time
    So if less energy is dissipated then why does the power rating increase
 
 
 
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